Two capillary tubes of same radius r but of l | vedclass

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(C) According to Poiseuille's law,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{P}{R}$,where $R = \frac{8\eta l}{\pi r^4}$ is t

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$\frac{l_1 l_2}{l_1 + l_2}$

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(C) According to Poiseuille's law,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{P}{R}$,where $R = \frac{8\eta l}{\pi r^4}$ is the fluid resistance.For tubes in parallel,the effective resistance $R_{eff}$ is given by $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$.Substituting the values of resistance: $\frac{\pi r^4}{8\eta l} = \frac{\pi r^4}{8\eta l_1} + \frac{\pi r^4}{8\eta l_2}$.Canceling the common terms $\frac{\pi r^4}{8\eta}$,we get $\frac{1}{l} = \frac{1}{l_1} + \frac{1}{l_2}$.Solving for $l$,we get $l = \frac{l_1 l_2}{l_1 + l_2}$.

Two capillary tubes of same radius $r$ but of lengths $l_1$ and $l_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the length of a single tube of the same radius $r$ that can replace the two tubes so that the rate of flow is same as before?

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